# Ch3-4 - Outline 3.1 Condence Intervals for Association...

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Chapter 3. Inference for Contingency Tables Deyuan Li School of Management, Fudan University Feb. 28, 2011 1/60 Outline 3.1 Confdence Intervals ±or Association Parameters 3.2 Testing Independence in Two-way Contingency Tables 3.3 Following-up Chi-squared Tests 3.4 Two-way Tables with Ordered Classifcations 3.5 Small-sample Tests o± Independence 3.6 Small-sample Confdence Intervals ±or 2 × 2Tab les 2/60 3.1 ConFdence Intervals for Association Parameters 3.1.1 Interval estimation of odds ratios For a 2 × 2 table, the sample odds ratio ˆ θ =( n 11 n 22 ) / ( n 12 n 21 ). I± any n ij =0 ˆ θ =0or . I± both entries in a row or column are zero ˆ θ is undefned. The expected value and variance o± ˆ θ and log ˆ θ do not exist. The Wald 100(1 α )% CI due to Woolf (1955). (1) calculate the estimated SE ±or log ˆ θ (derived in Section 3.1.7): { Insert . ..... } (2) construct the CI ±or log θ as log ˆ θ ± z α/ 2 ˆ σ (log ˆ θ ); (3) exponentiate the endpoints to obtain the CI ±or θ . Features: (1) a bit conservative; (2) the CI does not exist when ˆ θ . Other approaches (Section 3.1.8). 3/60 3.1.2 Aspirin and myocardial infarction example Table 3.1 Swedish Study on Aspirin Use and Myocardial In±arction. Myocardial In±arction Yes No Total Placebo 28 656 684 Aspirin 18 658 676 A randomized clinical trial, placebo vs. aspirin. Outcome = death due to myocardial in±arction within 3 years ˆ θ =1 . 56, log ˆ θ . 445 and ˆ σ (log ˆ θ )=0 . 307. A 95% CI ±or log θ is 0 . 445 ± 1 . 96 × 0 . 307 = ( 0 . 157 , 1 . 047). The 95% CI ±or θ is [exp( 0 . 157) , exp(1 . 047)] = (0 . 85 , 2 . 85). The width o± the CI is quite large. The estimate o± the true odds ratio is rather imprecise. The CI ±or θ contains 1.0. It is plausible that the true odds o± death are equal ±or aspirin and placebo. 4/60

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3.1.3 Interval estimation of diFerence of proportions Consider two independent binomial samples. For sample i with sample size n i , the probability of success is π i and the observed number of successes is y i . The Wald 100(1 α )% CI for π 1 π 2 . (1) calculate sample proportions ˆ π i = y i / n i ; (2) calculate the estimated SE for ˆ π 1 ˆ π 2 : { Insert . ..... } (3) construct the CI for π 1 π 2 as (ˆ π 1 ˆ π 2 ) ± z α/ 2 ˆ σ π 1 ˆ π 2 ). Such CI usually has true coverage probability less than the nominal con±dence coeﬃcient, especially when π 1 and π 2 are near 0 or 1. Better methods in Section 3.1.8. 5/60 For Table 3.1 example: Placebo: ˆ π 1 =28 / 684 = 0 . 0409, Aspirin: ˆ π 2 =18 / 676 = 0 . 0266. So, ˆ π 1 ˆ π 2 =0 . 0409 0 . 0266 = 0 . 014 , ˆ σ π 1 ˆ π 2 )=[ . 0409(1 . 0409) / 684 + . 0266(1 . 0266) / 676] 1 / 2 = . 0098 . The 95% CI for π 1 π 2 is 0 . 014 ± 1 . 96 × 0 . 0098 = ( 0 . 005 , 0 . 033) . The death rate for those taking placebo was between -0.005 and 0.033 larger than that for those taking aspirin. 6/60 3.1.4 Interval estimation of relative risk The Wald 100(1 α )% CI for r .
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## This note was uploaded on 11/20/2011 for the course ST 3241 taught by Professor Deyuanli during the Spring '11 term at Adams State University.

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Ch3-4 - Outline 3.1 Condence Intervals for Association...

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